Is the photon velocity regarding a point in a expanding universe constant?

Question

Could be a photon moving in an expanding universe be thought of as an ant moving along an elastic rope fixed at the upper end on a cieling that is streching due to a constant mass at its bottom end. So even the ant has a specific velocity regarding the elastic rope he could have different velocities regarding the end point hanging on the cieling because the points on the rope themselves are moving?

Answer 1

Could be a photon moving in an expanding universe be thought of as an ant moving along an elastic rope fixed at the upper end on a ceiling that is stretching due to a constant mass at its bottom end.

The point where the elastic rope meets with the ceiling can be thought of as an inertial point. I believe for this reason this analogy is not a good one to describe the speed of light. There's no general inertial frame in cosmology. All inertial frames are defined locally. In your analogy, it seems like the universe has a general inertial reference frame, which is not true.

I simply ask if two points with a comoving distance have a time for a photon to reach one from the other defined just as distance divided with c or the velocity is not just c but should be a function of space expansion...

Well you can describe the comoving distance ($d_c$) as a function of $t$, $a(t)$ or $z$. In order to find how much time took for light to travel between two comoving objects you can divide comoving distance by $c$.

The conformal time defined as

$$\eta = \frac{d_p}{c}$$

where $$d_p = c\int_{t_e}^{t_0} \frac{dt}{a(t)}$$

Thus $$\eta =\int_{t_e}^{t_0} \frac{dt}{a(t)}$$

PS: If you want to write the above equations in terms of $z$ you can do it like this

$$\eta=\int \frac {dt} {a}=\int\frac {da} {a\dot{a}}=\int\frac {da} {a^2H}$$ and we can write $$H(z)=H_0E(z)$$ $$E(z)=\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}$$

so we have,

$$\eta=\int\frac {da} {a^2H_0E(z)}$$

and $dz=-da/a^2$ so we can write,

$$\eta=-H_0^{-1}\int_z^0\frac {dz} {E(z)}$$

$$\eta=H_0^{-1}\int_0^{z}\frac {dz} {E(z)}$$

Answer 2

The analogy fails because the ant has different velocities relative to different observers, while according to Special Relativity photons always have the same velocity in free space, for all observers.

A closer analogy might be that the ant looks longer or shorter to different observers at different points on the elastic.